2 Answers
2

This does not answer the general question, but a simple special case, which I find cute:

Suppose $R$ is a finite dimensional algebra over a field of characteristic zero. Pick $x$, $y\in R$. Since $[x,y]$ is a commutator, its action on every finite dimensional $R$-module is of trace zero; on the other hand, since $[x,y]$ is by hypothesis idempotent, its trace on a module is the dimension of its image. It follows that $[x,y]$ acts by zero on every module so, because faithful modules do exist, it must be zero.

Later. Consider the non-commutative polynomial $f(x,y)=[x,y]^2-[x,y]$, and define new polynomial $h$ by $$h(x,y)=f(x,y)-f(y,x)$$ A simple computation shows that $h(x,y)=2[x,y]$.

If we plug elements of $R$ into $f$ we get zero, so the same is true of $h$. The above observation means then that $2[x,y]=0$ for all $x$, $y\in R$. If now $2$ is invertible in $R$, or at least not a divisor of zero, then we see that $R$ is commutative.

Proof from "Two elementary generalisations of boolean rings" AMM Feb.1986:
Let $Z(R):=\{x\!\in\!R;\: \forall y\!\in\!R\!: xy=yx\}$ denote the center of the ring. We notice that
\begin{equation*}
\label{eq:7.1}
xy-yx=(xy-yx)^2=(yx-xy)^2=yx-xy.
\tag*{(1)}
\end{equation*}
First we show a general lemma, that if $xy\!=\!0$ implies $yx\!=\!0$ $(\ast)$, then every idempotent $e$ is central. For arbitrary $r\!\in\!R$ we have $(e^2\!-\!e)r=0=e(er\!-\!r)$ and by $(\ast)$ we have $0=(er\!-\!r)e=ere\!-\!re$. Similarly, from $r(e^2\!-\!e)=0=(re\!-\!r)e$ by $(\ast)$ we get $0=e(re\!-\!r)=ere\!-\!er$. Therefore $re=ere=er$ which proves $e\!\in\!Z(R)$.

Next we prove, that all commutators are central. This follows from the previous paragraph, because the condition $(\ast)$ holds: if $xy=0$, then $yx$ $=yx-xy$ $=(yx\!-\!xy)^2$ $=(yx)^2$ $=y(xy)x$ $=0$. We have proved, that
\begin{equation*}
\label{eq:7.2}
\forall x,y\in R:\; xy-yx\in Z(R).
\tag*{(2)}
\end{equation*}

Next we prove that all squares are central. This we get from the equality $x(xy-yx)$ $\overset{(1)}{=}x(yx-xy)$ $\overset{(2)}{=}(yx-xy)x$, which tells us that $x^2y=yx^2$ , i.e.
\begin{equation*}
\label{eq:7.3}
\forall x\in R:\; x^2\in Z(R).
\tag*{(3)}
\end{equation*}

Next we prove, that $(xy)^2=(yx)^2$. This we achieve via $(3)$ by writing $yxy$ as a sum of squares:
$$(xy)^2 =x(yxy)=x\big((yx)^2+y^2-(yx\!-\!y)^2-y^2x\big)$$
\begin{equation*}
\label{eq:7.4}
\overset{(3)} {=}\big((yx)^2+y^2-(yx-y)^2-y^2x\big)x=(yxy)x=(yx)^2
\tag*{(4)}
\end{equation*}